2013 USAMO

Day 1

Problem 1

In triangle , points lie on sides respectively. Let , , denote the circumcircles of triangles , , , respectively. Given the fact that segment intersects , , again at respectively, prove that .

Solution

Problem 2

For a positive integer plot equally spaced points around a circle. Label one of them , and place a marker at . One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of distinct moves available; two from each point. Let count the number of ways to advance around the circle exactly twice, beginning and ending at , without repeating a move. Prove that for all .

Solution

Problem 3

Let be a positive integer. There are marks, each with a black side and a white side, arranged into an equilateral triangle, with the biggest row containing marks. Initially, each mark has the black side up. An operation is to choose a line parallel to the sides of the triangle, and flipping all the marks on that line. A configuration is called admissible if it can be obtained from the initial configuration by performing a finite number of operations. For each admissible configuration , let denote the smallest number of operations required to obtain from the initial configuration. Find the maximum value of , where varies over all admissible configurations.

Solution

Day 2

Problem 4

Find all real numbers satisfying

Solution

Problem 5

Given postive integers and , prove that there is a positive integer such that the numbers and have the same number of occurrences of each non-zero digit when written in base ten.

Solution

Problem 6

Let be a triangle. Find all points on segment satisfying the following property: If and are the intersections of line with the common external tangent lines of the circumcircles of triangles and , then

Solution

See Also

2013 USAMO (Problems • Resources)
Preceded by 2012 USAMO	Followed by 2014 USAMO
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions